Theorem ndmovg 7553

Description: The value of an operation outside its domain. (Contributed by NM, 28-Mar-2008.)

Assertion

Ref	Expression
ndmovg	⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Proof of Theorem ndmovg

Step	Hyp	Ref	Expression
1		df-ov 7373	. 2 ⊢ (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2		eleq2 2826	. . . . . 6 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆)))
3		opelxp 5670	. . . . . 6 ⊢ (⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆) ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆))
4	2, 3	bitrdi 287	. . . . 5 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
5	4	notbid 318	. . . 4 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
6		ndmfv 6876	. . . 4 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
7	5, 6	biimtrrdi 254	. . 3 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅))
8	7	imp 406	. 2 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
9	1, 8	eqtrid 2784	1 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1542   ∈ wcel 2114  ∅c0 4287  ⟨cop 4588   × cxp 5632  dom cdm 5634  ‘cfv 6502  (class class class)co 7370

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5245  ax-nul 5255  ax-pr 5381

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-mo 2540  df-eu 2570  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-br 5101  df-opab 5163  df-xp 5640  df-dm 5644  df-iota 6458  df-fv 6510  df-ov 7373

This theorem is referenced by:  ndmov  7554  curry1val  8059  curry2val  8063  1div0  11810  1div0OLD  11811  repsundef  14708  cshnz  14729  mamufacex  22357  mavmulsolcl  22512  mavmul0g  22514  iscau2  25250  1div0apr  30561  rrxsphere  49137